A circle with area $9\pi$ has a sector with a $\dfrac{1}{9}\pi$ radian central angle. What is the area of the sector? ${9\pi}$ $\color{#9D38BD}{\dfrac{1}{9}\pi}$ ${\dfrac{1}{2}\pi}$
Solution: The ratio between the sector's central angle $\theta$ and $2 \pi$ radians is equal to the ratio between the sector's area, $A_s$ , and the whole circle's area, $A_c$ $\dfrac{\theta}{2 \pi} = \dfrac{A_s}{A_c}$ $\dfrac{1}{9}\pi \div 2 \pi = \dfrac{A_s}{9\pi}$ $\dfrac{1}{18} = \dfrac{A_s}{9\pi}$ $\dfrac{1}{18} \times 9\pi = A_s$ $\dfrac{1}{2}\pi = A_s$